I've been learning about the Circle Method (at the level of the book
"An Invitation to Modern Number Theory," by Miller and Takloo-Bighash). The arguments in the book show how the
Circle method can be applied successfully to the ternary Goldbach problem, but fails for the binary problem (since the minor arc
contribution can't be bounded sufficiently well). Is this the final word on the Circle method's application to this problem? Is there any hope
that more sophisticated arguments/estimates might make it work?

Also, what book and/or set of notes would be good for me to continue studying the Circle Method? I have looked at
Vaughn's book, it is too dense for me, without enough motivating material. Is the book by Nathanson ("Additive Number Theory: The Classical Bases") a
good place to continue?

Thanks to everyone for the good suggestions, comments, and references. I should have made it clear in my question that I am not a professional mathematician looking to work on this, I'm a theoretical physicist with some spare time who wants to learn more about the Circle Method. I'm self taught in this area, which is why I need more motivational type books than Vaughn's . At some point I would like to try to get through Vaughn; I agree that hard work is the only way to make progress! Thanks, Tom
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Tom DickensMar 26 '11 at 0:52

As for applying the circle method to the binary Goldbach problem, my understanding is that the minor arc estimate is a very delicate matter (see, for example, the use of Vaughan's identity to get an upper bound on the value $F(\alpha) = \sum_{p \leq N} \log (p) e(p\alpha)$ on the minor arc. However, I believe a much more subtle point is the following quickly-glossed over fact of the proof: using Parseval's identity, we obtain the upper bound
$$\displaystyle \int_0^1 |F(\alpha)|^2 d\alpha = \sum_{p \leq N} (\log p)^2 \leq \log N \sum_{p \leq N} \log p \ll N \log N$$
This coupled with the upper bound for $|F(\alpha)|$ on the minor arc is what gives the integral estimate that allows one to conclude that the contribution of the minor arc to the integral is smaller than the contribution from the major arc. Note that the analogous bound for $\displaystyle \int_0^1 |F(\alpha)|d\alpha \ll N$ is much weaker, since one expects the contribution from the minor arc to be less than $\displaystyle \frac{N}{(\log N)^2}$. What made the ternary case work is that Parseval's identity allows a much better upper bound, and together with the estimate of $\displaystyle |F(\alpha)| \ll \frac{N}{(\log N)^{(B/2) - 4}}$ gives us a barely-good-enough bound on the minor arc contribution.

There are some concerns about the major arc estimates as well (and I believe this is a much more difficult obstacle than the minor arc, though this belief may be unfounded), but I am not able to come up with a lucid narrative at this point.

I really like the exposition in these notes of Soundararajan (start on page 18). His proof is conditional on GRH, which strips the method down to a very easily understood core. I think it would be extremely instructive to follow Sound's approach and attempt to prove binary Goldbach on GRH. You will fail, of course, but that's the point!

In addition, I really like the chapter on the circle method in Iwaniec and Kowalski. This is a more challenging read, but it very nicely illustrates the scope and applications of the method.

On a reference for the circle method: The book ``Analytic methods for diophantine equations and diophantine inequalities'' by Harold Davenport is superb. This is how I learned the circle method and some of its applications, and I highly recommend it.

I am a little nervous writing this, because it is far from my field. But it was my understanding that there is a much more basic problem in proving the binary Goldbach conjecture by analytic methods.

Let $P$ be any set of positive integers and let $f_P(x) = \sum_{p \in P} x^p$. Then the behavior of $f_P$ near $1$ depends on the rate of growth of $\pi_P(n) := \# \{ k \in P: \ k \leq n \}$. Similarly, the behavior of $f_P$ near $e^{2 \pi i \ell/m}$ depends on the rates of growth of the functions $\pi_P(n: r,m) := \# \{ k \in P : \ k\leq n,\ k \equiv r \mod m \}$. The cricle method is all about making this dependence precise, and about approximating $f_P(e^{i \theta})$ by $f_P(e^{2 \pi i \ell/m})$ for some rational approximation $\ell/m$ to $\theta/(2 \pi)$.

However, there is a set of integers $P$ for which all the $\pi_{P}(n: r,m)$ have the same growth rates as for the primes, yet binary Goldbach is false for $P$. So, rather than improving the details of the circle method, one is left with a fundamentally new problem: Thinking of new properties of the primes to make use of.

So, what is $P$? I'll first do it with $\pi_P(n)$ having the correct growth, then I'll add in the more refined $\pi_P(n, r,m)$'s.

Divide $\mathbb{Z}_{\geq 0}$ up into intervals $[a_0, a_1) \sqcup [a_1, a_2) \sqcup \cdots$ where $a_0=1$ and $a_{k+1} = a_k + \sqrt{a_k} + O(1)$ where the $O(1)$ is some absolute cnstant bound for how much error we will permit (I think $2$ would work). Choose a subsequence $b_j$ of the $a_i$ such that $b_1=a_1$ and $b_{i+1} \geq 2 b_i$. We will make sure that none of the $b_i$ are in $P+P$.

Don't put any elements of $[a_0, a_1)$ into $P$. That ensures that $b_1=a_1$ is not in $P+P$. We now inductively construct $[a_k, a_{k+1}) \cap P$. Suppose that we have already constructed the part of $P$ in $\bigcup_{i < k} [a_i, a_{i+1})$, and that $P+P$ does not contain any $b_j$. Let $b_j$ be the unique $b$ such that $b_{j-1} \leq a_k < a_{k+1} \leq b_j$. When adding elements of $[a_k, a_{k+1})$ to $P$, we can't make a sum of the form $b_{j'}$ with $j' < j$, because $b_{j'}< a_k$. And we can't make a sum $b_{j''}$ with $j' > j$, because $b_{j''} > 2 b_j \geq a_{k+1} + a_{k+1}$. So our only concern is that we might create the sum $b_j$.

Choose $\sqrt{a_k}/\log a_k+O(1)$ elements of $[a_k, a_{k+1})$ to put into $P$, subject to the sole condition that we don't make $b_j$ land in $P+P$. Again, the $O(1)$ is a once and for all global choice. In order to avoid creating this sum, we must avoid putting in $b_j - p$, where $p$ is an element of $P$ lying in an interval of length $[\sqrt{a_k}]$. The size of $P \cap [x, x+L]$ is $\approx L/\log(x+L) \leq L/\log L$. So we have to avoid $\sqrt{a_k}/\log \sqrt{a_k} = 2 \sqrt{a_k}/\log a_k$ things, while choosing $\sqrt{a_k}/\log a_k$ things, all from an interval of length $\sqrt{a_k}$. Since $\log a_k$ is eventually bigger than $3$, this is not a problem.

Now, let's do this while, at the same time, making sure that the $\pi_P(n, r,m)$ grow correctly. This time, make sure to take the $b_j$ to be even, since you won't be impressed if they are odd. The $O(1)$ in the defining recursion for the $a$'s gives me enough room to make sure lots of $a$'s are even, so that's fine. This time, we are going to require that, in addition, for every $m$, we have $(\sqrt{a_k}/\log a_k)/\phi(m!)+O(1)$ elements of $P \cap [a_k, a_{k+1})$ in each invertible residue class modulo $m!$. Note that, for $m$ large, the $O(1)$ swamps the first term, so this condition is trivial. So for each $k$, this is in fact finitely many conditions to obey.

And, since $\log a_k$ does eventually get larger than $3 \phi(m!)$, this should be doable.

There is a lot of unchecked material here; I welcome corrections and citations.

Dear David, I would argue with the "circle method is all about making this dependence precise". In most applications of the method there is a distinction between the major and the minor arcs. Your observations (that I admittedly have not studied yet) touch upon the major arcs. Now, eventually, the major arcs do produce the heuristic Hardy-Littlewood count when they are properly defined. So the minor arcs do produce a negligible contribution if you believe Hardy-Littlewood. The problem is that we are unable to estimate the integral on the minor arcs (in the case of the binary Goldbach problem).
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GH from MOMar 25 '11 at 16:43

If you want to do research in the circle method, then my best advice is to study Vaughan's book as it is the definite source on the subject. I have studied it by myself (as an undergraduate) and soon it helped me to write my first papers (using the circle method). I also found the circle method to be of great use in my thesis work. You can only grow strong if you train yourself on difficult material (reading and problems). I am not saying this to be arrogant, but because this is what I was told and this is what I experienced. I make my students suffer, and it serves them well.

As to your original question: you can never say something does not work until you have proof that it is the case (I learned this wisdom from Peter Sarnak). I think one main difficulty with applying the circle method to the binary Goldbach problem is the fact that the $L^p$-norms of the generating function are too large for $p<2$. Unfortunately, I don't know a reference off-hand. Note also that the circle method is capable of showing that up to a large $x$ all but $x^{0.99}$ even numbers are a sum of two primes, and there has been recent progress on decreasing the exponent here.

"You should study Vaughan's book. If it is too dense for you, then the circle method is not your piece of cake." -- I totally disagree. I have nothing bad to say about Vaughan's book, but if the OP finds it difficult then I think he is absolutely right to look for a gentler reference.
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Frank ThorneMar 25 '11 at 7:17

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+1 as I find parts of your answer interesting. However, the description on the page of the questioner suggest to me that he is essentially studying this material completely on his own. He might thus be in a very different situation than from the one in which you were when learning this material. Also, different people are different. If one is 'alone' sources giving motivation and intuition can be crucial; if one has other ways to pick up this, often almost unwittingly, (via talks, seing how people in the know talk/think about/work with the material), then not.
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quidMar 25 '11 at 14:58

Thanks, unknown. You and Frank Thorne made me refine my response. I am not sure it will make more people happy though.
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GH from MOMar 25 '11 at 15:23